find the condition that `x^3 -px^2 +qx -r=0` may have two roots equal in magniude but of opposite sign

Answer

To find the condition that the equation \( x^3 - px^2 + qx - r = 0 \) may have two roots equal in magnitude but of opposite sign, we can follow these steps:

### Step 1: Understand the Roots
Let the roots of the equation be \( \alpha, -\alpha, \) and \( \beta \). Here, \( \alpha \) and \( -\alpha \) are the roots that are equal in magnitude but opposite in sign.

### Step 2: Use Vieta's Formulas
According to Vieta's formulas, for a cubic equation \( ax^3 + bx^2 + cx + d = 0 \):
- The sum of the roots \( \alpha + (-\alpha) + \beta = -\frac{b}{a} \)
- The sum of the product of the roots taken two at a time \( \alpha(-\alpha) + \alpha\beta + (-\alpha)\beta = \frac{c}{a} \)
- The product of the roots \( \alpha \cdot (-\alpha) \cdot \beta = -\frac{d}{a} \)

### Step 3: Apply Vieta's Formulas
For our equation \( x^3 - px^2 + qx - r = 0 \):
- Here, \( a = 1, b = -p, c = q, d = -r \).

Using Vieta's formulas:
1. The sum of the roots:
\[ \alpha - \alpha + \beta = p \implies \beta = p \]

2. The sum of the product of the roots taken two at a time:
\[ \alpha(-\alpha) + \alpha\beta + (-\alpha)\beta = q \]
Substituting \( \beta = p \):
\[ -\alpha^2 + \alpha p - \alpha p = q \implies -\alpha^2 = q \implies \alpha^2 = -q \]

3. The product of the roots:
\[ \alpha \cdot (-\alpha) \cdot \beta = -r \implies -\alpha^2 \cdot p = -r \implies \alpha^2 p = r \]

### Step 4: Relate the Conditions
From the equations derived:
1. \( \alpha^2 = -q \)
2. \( \alpha^2 p = r \)

### Step 5: Substitute and Find the Condition
Substituting \( \alpha^2 \) from the first equation into the second:
\[ -q \cdot p = r \implies pq + r = 0 \]

### Conclusion
Thus, the condition that the equation \( x^3 - px^2 + qx - r = 0 \) may have two roots equal in magnitude but of opposite sign is:
\[ pq - r = 0 \]